Solve a^2+3/2+63/2a=0

$${ a }^{ 2 } + \frac{ 3 }{ 2 } + \frac{ 63 }{ 2a } = 0$$

Solve for a

$$a=-3$$

Solution Steps

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a, the least common multiple of 2,2a.

$$2aa^{2}+2a\times \frac{3}{2}+63=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$2a^{3}+2a\times \frac{3}{2}+63=0$$

Multiply 2 and \frac{3}{2} to get 3.

$$2a^{3}+3a+63=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 63 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{63}{2},±63,±\frac{21}{2},±21,±\frac{9}{2},±9,±\frac{7}{2},±7,±\frac{3}{2},±3,±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=-3$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide 2a^{3}+3a+63 by a+3 to get 2a^{2}-6a+21. Solve the equation where the result equals to 0.

$$2a^{2}-6a+21=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, -6 for b, and 21 for c in the quadratic formula.

$$a=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 21}}{2\times 2}$$

Do the calculations.

$$a=\frac{6±\sqrt{-132}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$a\in \emptyset$$

List all found solutions.

$$a=-3$$